I’ve bet against three of my friends that I am correct on this one.
On a completely level plane (no earth curvature) if a gun is fired parallel to the ground from say 5 feet up, and at the moment the bullet leaves the muzzle a bullet is dropped from 5 feet up,
they will hit the ground at the same time.
I contend this is true since there is no force counteracting gravity. The perpendicular force given to the bullet by the gun has no effect on gravity.
They say “no way” the fired bullet will stay up longer.
We’re assuming this is a plane surface in lieu of the Earth, with the gravity acting in parallel at all points on the plane surface. However, if we are talking about a plane surface on the Earth, then the dropped bullet will hit first, because the travelling bullet will get farther from the Earth’s center than the stationary one, and be affected just slightly less, the farther it goes. I know I’m explaining it badly, but I can’t be bothered to do calculations right now.
Spin would let the bullet go further horizontally since it reduces drag, but even if the bullet was aerodynamically perfect (or fired through a vaccuum, which makes spin irrelevant) it would still drop at the same rate.
If the gun barrel is four feet up and the bullet is fired horizontally with any amound of force or dropped, it’ll hit the ground in half a second and I can’t see any way to change that.
Well, to more precisely explain my response to the OP, I was talking about small arms “guns.” I suppose big-axe guns like stuff on battleships would require complex calculations and explanations, like Q.E.D. suggests.
I don’t think spin would have an effect. Longitudinal spin gives the bullet stability, but it wouldn’t create any vertical pressure difference. Compare to a spin around the pitch axis (backspin or forespin) which could create a lift force.
As long as the bullet is symmetrical, air resistance will also have no effect. If the bullet were assymetrical or had an angle of attack to its direction, it could generate a lift force, but drag alone (due to air resistance on a symmetrical body) will affect only the horizonal component, not the vertical.
Spin doesn’t enter into the problem unless the fired bullet hit somebody. Then you’d need a spin doctor to say, “Bullets don’t kill people, science experiments kill people.”
So, the drag acts against the direction of travel, and when the fired bullet is arcing downward, that drag slows its descent as the vertical component of the drag (same ratio as the vertical component of the instantaneous velocity) acts “upward”. The drag being a function of the square of the velocity tells me the total upward component of drag for the fired bullet will be greater than the total upward drag for the dropped bullet.
Uncle Bill, I need to think about your cite some more but I don’t think it’s correct. I don’t think it’s properly treating drag as a vector. Yes, the drag on the fired bullet is much higher, but this drag force is a vector directed substantially horizontal. Only a very small component of that drag would be directed vertically, and that component would be equal to the vertical drag on the dropped bullet because they have identical vertical velocity components. The horizontal velocity can’t contribute to drag in the vertical direction.
I know I’ve made a circular argument above, saying that the vertical velocities of the bullets must be the same because the drag is the same because the velocity is the same. The point is that (in the absence of aerodynamic effects like lift) the only force acting on the two bullets in the vertical direction is gravity. This force acts identically on both bullets so it imparts the same vertical acceleration, resulting in the same vertical velocity. This velocity results in another vertical force, drag, but it is the same for both bullets because they have the same velocity. The fact that the fired bullet has a huge horizontal velocity and, hence, a huge horizontal drag force, has no effect whatsoever on the vertical forces or velocities. That’s kind of the point of the exercise, demonstrating that the components of these equations in two orthogonal directions can be decoupled.
The fact that the bullet arcs downward is due to the vertical velocity component caused by the gravitational force which is summed with the horizontal component to create a vector pointing at an angle below horizontal. The arcing downward is not caused by any part of the horizontal velocity switching to vertical velocity (and thereby contributing to some additional vertical drag) because that would violate Newton’s laws of motion.
I may be wrong in saying the two bullets hit the ground at the same time, but that answer is certainly true for first order effects. In any case, I think the drag argument in your cite is incorrect.
If you fired a bullet downwards, it would penetrate the floor/ground, but if you were to simply drop a bullet, it wouldnt penetrate the floor/ground. Right?? If this is true, then how can it be they are traveling at equal speeds?
They aren’t travelling at equal speeds, but they have equal vertical velocities (or not, which is what we’re discussing). Velocity is a vector, which means it has magnitude (speed) and direction. If you fire a gun horizontally, the bullet has a very high horizontal velocity and acquires a relatively small vertical velocity due to the force of gravity. The bullet is still moving very fast, but it’s not moving down very fast.
I second micco’s point that the drag has to be separated into horizontal and vertical components, and that UncleBill is therefore wrong.
Another factor that most of you have missed: If the bullet is not spherical, the dropped bullet will experience different amounts of drag depending on the position it is in while falling. Pointing up, it will fall slower than pointing down. And I’d guess that if the length is significantly greater than the width, then pointing sideways will be slower still.
If I read the cite right, then it can, because drag is proportional to the square of the velocity. (However, I do think the cite is hard to understand.) Take an example.
Say drag force is equal to DF = cV[sup]2[/sup], where c is some constant and V is the velocity, and it acts in the opposite direction of the V vector.
Bullet A has (V[sub]X[/sub], V[sub]Y[/sub]) = (0, -1) in whatever units. Thus V[sup]2[/sup] = 1 and so DF has a magnitude equal to 1c. Thus for Bullet A, DF = (0, 1c).
Bullet B has (V[sub]X[/sub], V[sub]Y[/sub]) = (100, -1), and so V[sup]2[/sup] = 10001, and so DF has magnitude 10001c. Thus for Bullet B, DF = (10000.500c, 100.005c). As you can see, the bolded Y-components are not the same.
But wouldn’t it be proper to separate the horizonal drag (which will decrease the horizontal velocity so the bullet will not be able to go as far horizontally before reaching the ground) from the vertical drag (which will decrease the vertical velocity so the bullet will take longer to reach the ground)?
And the vertical drag would be equal for both bullets, no?
